Properties of Two Events and Independence Test

To determine which of the given options are correct, we use the definitions of conditional probability and basic probability rules:

1. Calculate $P(Y)$: By the definition of conditional probability: $P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)}$ Given $P(X \mid Y) = \frac{1}{2}$ and $P(X \cap Y) = \frac{1}{6}$, we have: $\frac{1}{2} = \frac{1/6}{P(Y)} \implies P(Y) = \frac{1/6}{1/2} = \frac{1}{3}$

2. Calculate $P(X)$: Similarly: $P(Y \mid X) = \frac{P(X \cap Y)}{P(X)}$ Given $P(Y \mid X) = \frac{1}{3}$ and $P(X \cap Y) = \frac{1}{6}$, we have: $\frac{1}{3} = \frac{1/6}{P(X)} \implies P(X) = \frac{1/6}{1/3} = \frac{1}{2}$

3. Check for Independence (Options B and C): Two events $X$ and $Y$ are independent if: $P(X \cap Y) = P(X) \cdot P(Y)$ Evaluating the product: $P(X) \cdot P(Y) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$ Since $P(X \cap Y) = \frac{1}{6} = P(X) \cdot P(Y)$, the events $X$ and $Y$ are independent. Therefore, Option (B) is correct and Option (C) is incorrect.

4. Calculate $P(X \cup Y)$ (Option A): Using the addition rule for probabilities: $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$ $P(X \cup Y) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6} = \frac{3 + 2 - 1}{6} = \frac{4}{6} = \frac{2}{3}$ Therefore, Option (A) is correct.

5. Calculate $P(X^C \cap Y)$ (Option D): We have: $P(X^C \cap Y) = P(Y) - P(X \cap Y) = \frac{1}{3} - \frac{1}{6} = \frac{1}{6}$ Option (D) claims $P(X^C \cap Y) = \frac{1}{3}$, which is incorrect.

Thus, the correct options are A and B.